Proving Horizontal Component of Tension in Stationary Chain

[RoyalCat]

I'd just like some confirmation on my answers, and I'd appreciate it if someone could take the time to explain why what I did is right. I solved it with a lot of hand-waving, so I'm very unsure of how I reached the answers.

A uniform chain of weight $$W$$ is strung between two vertical walls. The chain makes an angle $$\theta$$ below the horizontal at points A and B, both of which are of equal height, one on each wall.

The lowest point of the chain is denoted as point $$P$$

A. Prove that the horizontal component of the tension in the chain is the same along every point in the chain.
B. Calculate the tension in the chain at the points A and B.
C. Consider the extreme cases where:
i. $$\theta \rightarrow 0$$
ii. $$\theta \rightarrow \tfrac{\pi}{2}$$

My answers:

I chose to look at the chain as a uniform rope. We weren't told anything about its length or internal structure, so it seems like a reasonable way to treat it.

A. Let's consider a small segment of the chain (Not an infinitesimal one). Gravity acts through its center of mass, there is a tension outwards (For the sake of discussion, up and right) from the top of the segment, and a tension going down and left from the bottom of the segment.
The only forces acting horizontally are the horizontal components of the two tension forces, and since the segment of the chain is at rest, then these two components must be equal.
As we chose a nondescript section of the chain, with the tensions acting on different points on the chain, from this we can conclude that the horizontal component of the tension is uniform across the chain.

B. Symmetry allows us to refer to just half the chain. We know the vertical (Thanks Mapes!) component of the tension at point A has to be equal to $$\tfrac{mg}{2}$$
$$T\sin{\theta}=\tfrac{mg}{2}$$
$$T_A=T_B=\tfrac{mg}{2\sin{\theta}}$$

C.
i. $$\theta \rightarrow 0$$
The tension explodes. You would need infinite tension force to keep the chain that taught.
This makes perfect sense, since a purely horizontal force cannot counter-balance a vertical one.
ii. $$\theta \rightarrow \tfrac{\pi}{2}$$
The limit is $$\tfrac{mg}{2}$$
The two walls would have to be touching, but then A and B each would just support half the weight of the chain. Practically, we're just looping the chain and connecting both ends to one point.

I've got this irking sensation that my thinking is flawed somewhere.

 

[Mapes]

I like your reasoning on part A.

We know the horizontal component of the tension at point A has to be equal to $$\tfrac{mg}{2}$$

You mean "vertical" here, right?

Everything else looks good to me.

 

[RoyalCat]

I like your reasoning on part A.



You mean "vertical" here, right?

Everything else looks good to me.

Whoops, that is correct. Thanks for catching it and for reviewing my work. :)